I have some questions about the Kitaev toy model for Majorana fermions (arXiv:cond-mat/0010440). First of all, his proof for the definition of the 'Majorana number' is not so clear to me.

$$P(H(L_{1} + L_{2})) = M(H)P(H(L_{1}))P(H(L_{2}))$$

where $P(H(L))$ denotes the ground state parity of a Hamiltonian $H(L)$ of a closed chain. Now, I want to proof if $M(H) = -1$, there are Majorana operators $\gamma_{1}$, $\gamma_{2L}$ associated with the ends of the chain.

The idea by Kitaev is, that we have two ground states

$$-i\gamma_{1} \gamma_{2L}|\Psi_{0}\rangle = |\Psi_{0}\rangle,\; -i\gamma_{1} \gamma_{2L} |\Psi_{1}\rangle = -|\Psi_{1}\rangle$$

where $|\Psi_{0}\rangle$ has an even fermionic parity and $|\Psi_{1}\rangle$ has an odd parity (these two cases represent the two phases; but the question is which states represent which case? My opinion is that the ground state with the odd parity represent the non-trivial case).

In the next step, Kitaev defines a effective parity operator $P = s(L)(-i\gamma_{1} \gamma_{2L}) = s(L)(-1)^{\alpha}$, where $s(L) = \pm 1$ (in which case is $s(L) = 1$ and $s(L) = -1$?) and $\alpha \in \{0,1\}$ for the state $|\Psi_{\alpha}\rangle$. The formal definition of the parity operator is

$$P = \prod_{j = 1}^{L}-i\gamma_{2j - 1}\gamma_{2j} = -i\gamma_{1}\prod_{j = 1}^{L-1}(-i\gamma_{2j}\gamma_{2j+1})\gamma_{2L},$$

but I don't know what is the reason that $\prod_{j = 1}^{L-1}(-i\gamma_{2j}\gamma_{2j+1}) = 1$, so that I get the definition of the effective parity?

For the proof with $M = -1$ Kitaev considers two chains, one of length $L_{1}$, the other of length $L_{2}$. For a closed chain the effective Hamitlonian is $H(L) = (i/2) u\gamma_{1}\gamma_{2L}$. The parameter $u$ represents direct interaction between the chain ends. The ground state of the closed chain is $|\Psi_{1}\rangle$ if $u > 0$, and $|\Psi_{0}\rangle$ if $u < 0$. Hence, $P(H(L)) = -s(L) \rm{sgn}(u)$. There are two ways to close the chains (see Fig. 3 in the publication by Kitaev). Both cases can be described by effective Hamiltonian

$$H(L_{1})H(L_{2}) = \frac{i}{2}u(\gamma_{1}^{\prime\prime}\gamma_{1}^{\prime} + \gamma_{2}^{\prime\prime}\gamma_{2}^{\prime})$$

$$H(L_{1} + L_{2}) = \frac{i}{2}u(\gamma_{1}^{\prime\prime}\gamma_{2}^{\prime} + \gamma_{2}^{\prime\prime}\gamma_{1}^{\prime})$$

It follows that

$$P(H(L_{1}))P(H(L_{2})) = s(L_{1})s(L_{2})$$

$$P(H(L_{1} + L_{2})) = -s(L_{1})s(L_{2})$$

Because of this definition of the Majorana number holds for $M = -1$ but I don't seen that it holds since I don't know how $s(L)$ is defined!

Kitaev shown that for a quadratic Hamiltonian

$$H = \frac{i}{4}\sum_{l,m}A_{lm}\gamma_{l}\gamma_{m}$$

the parity operator is

$$P(H) = \rm{sgn}(Pf(A))$$

where $\rm{Pf}(A)$ is the pfaffian of matrix $A$. (How can I show this? I have no idea!)

Why can I write $M(H) = \rm{sgn}(\rm{Pf}(A))$? Further what is the $\rm{sgn}$ of a Pfaffian (the Definition of $\rm{Pf}(A)^{2} = \rm{det}(A)$)?

  • $\begingroup$ Definition of the Pfaffian (including the sign): en.wikipedia.org/wiki/Pfaffian (First equation in the Section "Formal definition".) You get Pfaffians when you normal order fermionic operators. $\endgroup$ – Norbert Schuch Aug 22 '13 at 23:15
  • $\begingroup$ Yes, that is correct. However, a different idea is that $A_{lm}$ is skrew-symmetric in Majorana basis. A skrew-symmetric matrix can with the tranformation $WAW^{T}$ in Jordan normal form. I know that $\rm{sgn}(Pf(WAW^{T}) = \rm{sgn}(\rm{det}(W)\rm{Pf}(A))$ but Kitaev write that $\rm{sgn}(Pf(A) = \rm{sgn}(\rm{det}(W)) = \rm{sgn}(\rm{Pf}(A))$ and I don't have any idea how I can proof it. Further, I don't know how I combined it with the parity operator. $\endgroup$ – user27964 Aug 23 '13 at 4:44
  • $\begingroup$ Okay, it is easy to show that $\rm{sgn}(\rm{det}(W)) = \rm{sgn}(\rm{Pf}(A))$ with the properties of a skrew-symmetric matrix. However, I don't see why the sign of the determinant of the ortogonal matrix shows that the parity is changing? $\endgroup$ – user27964 Aug 23 '13 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy